Miscellaneous · Q16 (MCQ): When is $\vec a\cdot\vec b\ge0$⁠?

Basics you need

Solution

1. The sign of $\vec a\cdot\vec b$ is the sign of $\cos\theta$

   $$\vec a\cdot\vec b = \underbrace{|\vec a||\vec b|}_{> 0}\cos\theta.$$

   So $\vec a\cdot\vec b \ge 0$ if and only if $\cos\theta \ge 0$⁠.

   The question speaks of the angle between $\vec a$ and $\vec b$⁠, and an angle is defined only when both vectors are nonzero. Then $|\vec a||\vec b|$ is a positive number, and multiplying by a positive number does not change a sign.
   Section 10.6.1, Definition 2. Lesson: the dot product

2. Where $\cos\theta \ge 0$ on $[0, \pi]$

   $$\cos\theta \ge 0,\ 0 \le \theta \le \pi \iff 0 \le \theta \le \frac{\pi}{2}.$$
   The cosine decreases from 1 at $\theta = 0$ to 0 at $\theta = \frac{\pi}{2}$ and on to $-1$ at $\theta = \pi$⁠. Both ends of $\left[0, \frac{\pi}{2}\right]$ belong to the set: at $\theta = 0$⁠, $\vec a\cdot\vec b = |\vec a||\vec b| > 0$⁠, and at $\theta = \frac{\pi}{2}$⁠, $\vec a\cdot\vec b = 0$⁠, which still satisfies "$\ge 0$".
   Section 10.6.1, Observations 2 and 3. Basics: cos θ is a shadow

3. Examine every option

   (A) $0 < \theta < \frac{\pi}{2}$⁠: leaves out $\theta = 0$ (parallel vectors, $\vec a\cdot\vec b = |\vec a||\vec b| > 0$⁠) and $\theta = \frac{\pi}{2}$ (perpendicular vectors, $\vec a\cdot\vec b = 0$⁠). At these angles $\vec a\cdot\vec b \ge 0$ holds, so (A) is not the exact condition.

   (B) $0 \le \theta \le \frac{\pi}{2}$⁠: exactly the set found in Step 2. (B) is correct.

   (C) $0 < \theta < \pi$⁠: contains $\theta = \frac{2\pi}{3}$⁠, where $\vec a\cdot\vec b = -\frac12|\vec a||\vec b| < 0$⁠, and leaves out $\theta = 0$⁠. (C) fails.

   (D) $0 \le \theta \le \pi$⁠: every possible angle, including $\theta = \pi$⁠, where $\vec a\cdot\vec b = -|\vec a||\vec b| < 0$⁠. It is true of every pair of vectors, so it says nothing about the sign. (D) is not the exact condition.

   The intended reading is the exact condition: the set must contain every angle where $\vec a\cdot\vec b \ge 0$ and no angle where it fails. (A) misses two angles where the inequality holds, while (C) and (D) contain angles where it fails. Read word for word, "only when" asks only for the first half (every angle with $\vec a\cdot\vec b \ge 0$ lies in the set). On that reading (A) and (C) still fail, because each leaves out $\theta = 0$⁠, but (D) passes trivially, since every angle lies in $[0, \pi]$⁠. An option that holds for all vectors gives no information, so (B) is the answer intended, and it is the one the key gives.
   Section 10.6.1, Observations 2, 3 and 4. Lesson: the multiple-choice questions

4. Check with vectors

   Take $\vec a = \hat i$ and three choices of $\vec b$⁠:

   $$\begin{aligned} \vec b = \hat i + \hat j&:\ \theta = \tfrac{\pi}{4},\ \vec a\cdot\vec b = 1 > 0, \\ \vec b = \hat j&:\ \theta = \tfrac{\pi}{2},\ \vec a\cdot\vec b = 0, \\ \vec b = -\hat i + \hat j&:\ \theta = \tfrac{3\pi}{4},\ \vec a\cdot\vec b = -1 < 0. \end{aligned}$$
   The first two angles lie in $\left[0, \frac{\pi}{2}\right]$ and give $\vec a\cdot\vec b \ge 0$⁠; the third lies outside and gives a negative product, as (B) predicts.
   Scalar product in components (Section 10.6.1). Lesson: the dot product

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